Comments by "EebstertheGreat" (@EebstertheGreat) on "A Problem with Rectangles - Numberphile" video.

  1. The "breaking down" and "building up" methods both always work (even for irrational rectangles), adding 3 and 4 rectangles, respectively. The only exception is that building up is not possible with squares. That means if it is possible to construct a tiling with n non-square rectangles, it is always possible to do so with n+3, n+4, n+6, n+7, and n+8 rectangles, and therefore any greater number. This leaves special cases of n+1, n+2, and n+5. In the example from this video, the 2:1 ratio gives a simple way to solve n=2, and therefore the special cases were n=3, n=4, and n=7. Additionally, after finding an n that works, we have to check every m < n to see if that one works. In the case of the 2:1 rectangles, that was just n=1. n=1 will only ever work for squares. n=2 will only ever work for 2:1 rectangles. n=3 will only work for 3:1 or 3:2 rectangles. This makes 1:1, 2:1, 3:1, and 3:2 the best ratios to use. 2:1 was in the video, so let's look at 1:1 (squares). Trivially, n=1 is already solved. n=2 is clearly impossible, as is n=3. n=4, 7, ... work by breaking down n=1. The special case of n=5 can be shown to be impossible, though the proof isn't easy. n=6 is possible, putting one square with side 2/3 in a corner and five more squares with side 1/3 around it. Breaking that down gives n=9.The next special case is n=8, which we solve in a similar way. Put a square with side 3/4 in a corner and surround it with seven squares with side 1/4. So we can do n=6, 7, and 8, and we can break down to add 3, therefore we can do every n ≥ 6. In summary, you can tile a square using any number of squares of any size except n=2, 3, or 5. For other rectangles, we can both break down (n↦n+3) and build up (n↦n+4). However, n=1, 2, 3 will never work (except for ratios 1:1, 2:1, and either 3:1 or 3:2, respectively). For each real number x, where the shorter side is x times the length of the longer side, we have to analyze that rectangle and find the least n that tiles the larger rectangle. There is an easy case that works for all rational x. If we have an a:b rectangle, with a, b integers, then we can always tile b*a of them into a square. So in the worst case, n=ab will always work. This means the special cases to test for each rectangle are 3 < n < ab and n = ab+1, ab+2, and ab+5. For instance, a 3:2 rectangle can tile a square by choosing a single size of 1/2 by 1/3 and tiling six of them in the square in a 3x2 grid. Thus for 3:2, we know n=6 works. But we can actually do better, since n=3 actually works. One rectangle has dimensions 1x2/3, and the other two dimensions 1/2x1/3. The special cases are then n=4, 5, and 8. It's easy to see that n=4 does not work. Whether n=5 or n=8 work are left as an exercise for the reader. Irrational x can also work. For instance, rectangles can solve n=6, by putting one vertically against one side and the other five horizontally against the other side. In this case, the larger rectangle has height 1 and width x, while the smaller rectangles have height (1-x) and width 1/5. Thus 1/x = (1-x)/(1/5), and using the quadratic formula gives two solutions: x = 1/2 ± √(5)/10. I don't know if other types of irrational numbers like constructible, cubic, or even transcendental numbers can work, but I somewhat doubt it.
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  2.  @dylanwinestone4625  You said you came up with the n+4 method from the start, then realized you hadn't come up with the n+4 method. I don't understand.
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  3.  @leif1075  The point is that the rectangles are all similar. They can be of different sizes but their sides have to be in the same ratio. In the example in the video, they were all 2:1.
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  4. ​ @zanti4132  I don't see it. The long rectangle has dimensions 1 : 1/k, and the shorter rectangles have ⅓k : ⅓. Thus k/3 + 1/k = 1. Multiplying by 3k gives the polynomial equation k² - 3k + 3 = 0. This has no real roots. In general, this setup has n+1 rectangles: one larger and n smaller. The larger rectangle has dimensions 1 : 1/k and the shorter ones have dimensions k/n : 1/n. Thus in general, k/n + 1/k = 1. So we always get a quadratic equation k² - nk + n = 0. This has real roots whenever n ≥ 4. In the n=4 case, we get k=2, which was shown in the video. I gave the n=5 case. The next one is n=6, with solutions k = 3±√3. In general, the two solutions are (n±√(n²-4n))/2,.
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  5.  @zanti4132  Oh right of course, I'm not sure what I was thinking. I was just misunderstanding your solution. In fact, on a bit of reflection, I think we could find algebraic solutions of nth order for any n using at least n different sizes of rectangles. I mean, it comes right out of the equations. By the same vein, if we are restricted to using finitely many rectangles, I am pretty sure any solution will have an algebraic ratio. Because you can always just construct a system of polynomial equations for any solution. However, by using infinitely many rectangles, we should be able to find transcendental solutions like log 2.
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  6. You can only divide a square into 2 rectangles with sides in the ratio 2:1, and vice-versa.
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  7.  @leif1075  How do you propose dividing the square into two similar rectangles that don't have sides in a 2:1 ratio?
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